= Solution
Use $t\geq0$ to label the initial half-line, with $(x,y,z)=(t,-t,1)$ at $s=0$. In the general characteristic from part (i), this sets $a=t$, $b=-t$, $c=1$. The parametric <solution surface> is therefore
$$
\boxed{x=t+s,\qquad y=-t+s+s^2,\qquad z=1+2s,\qquad t\geq0.}
$$
At $s=0$ this recovers exactly the given data. Moreover $(x_s,y_s,z_s)=(1,1+2s,2)=(1,z,2)$, so the surface follows the required characteristic <vector> field. The branch defining a classical graph through the data is determined below by restricting to the side of the projection fold containing $s=0$.
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